Hello Rozana
$\displaystyle Q(x,y,z)$ is an example of a propositional function; that is, a function that returns a truth value (i.e. the value True or False) that will depend upon the value(s) of any parameter(s) that are supplied to it.
Here, the value of $\displaystyle Q$ will be determined by the values of three parameters, $\displaystyle x, y$ and $\displaystyle z$. The values that $\displaystyle x, y, z$ can take are given in the question - defined as the domain of $\displaystyle Q(x,y,z)$.
Since $\displaystyle x$ can take $\displaystyle 3$ different values $\displaystyle (0,1,2) $; and $\displaystyle y$ can take $\displaystyle 2$ different values $\displaystyle (0,1)$; and $\displaystyle z$ can also take $\displaystyle 2$ values $\displaystyle (0,1)$, there are $\displaystyle 3\times2\times2= 12$ possible propositions that can be represented by $\displaystyle Q(x,y,z)$, each one returning the value True or False. They are:$\displaystyle Q(0,0,0),\; Q(0,0,1),\; Q(0,1,0),\;Q(0,1,1)$
$\displaystyle Q(1,0,0),\; Q(1,0,1),\; Q(1,1,0),\;Q(1,1,1)$
$\displaystyle Q(2,0,0),\; Q(2,0,1),\; Q(2,1,0),\;Q(2,1,1)$
Without knowing the actual details of what $\displaystyle Q$ is, we can't say what value each of these propositions has. However, we can say, that if (as in your original question) $\displaystyle \exists z,\; \neg Q(0,0,z)$ is a true statement, then for one or both of the possible values of $\displaystyle z$ ($\displaystyle 0$ and $\displaystyle 1$), $\displaystyle \neg Q(0,0,z)$ is true. In other words, $\displaystyle \neg Q(0,0,0)$ is true or $\displaystyle \neg Q(0,0,1)$ is true (or both). So:$\displaystyle \exists z,\; \neg Q(0,0,z)$ can be written as $\displaystyle \neg Q(0,0,0) \lor \neg Q(0,0,1)$
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Question 3 can be answered in the same way:$\displaystyle \forall y,\;Q(0,y,0)$
meansFor all $\displaystyle y$ (that is, $\displaystyle y = 0$ and $\displaystyle y = 1$), $\displaystyle Q(0,y,0)$ is true.
So$\displaystyle Q(0,0,0)$ is true and $\displaystyle Q(0,1,0)$ is true.
Therefore:$\displaystyle \forall y,\;Q(0,y,0)$ can be written as $\displaystyle Q(0,0,0) \land Q(0,1,0)$
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Question 4
Examine the truth of each of the propositions in turn:$\displaystyle p:\forall x, \; x+1>x$
$\displaystyle p$ is True. It simply says that if we add $\displaystyle 1$ to any real number, the answer is greater than the number we started with.$\displaystyle q:\forall x, \;2x = 3x$
Clearly $\displaystyle q$ is False. To show this, we only need find one value of $\displaystyle x$ for which $\displaystyle 2x \ne 3x$. (That's not very hard, is it?)$\displaystyle r:\exists x,\;x=-x$
$\displaystyle r$ is True. There is a value of $\displaystyle x$ for which $\displaystyle x = -x$; that is $\displaystyle x = 0$.
So the ordered triple $\displaystyle (p,q,r)$ has the truth value $\displaystyle (T, F, T)$.
Does that clear up the problems?
Grandad